Question

If $A=\left[\begin{array}{cc}1& 0\\ -1& 7\end{array}\right]$ and $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right],$ then find the $k$ so that $A^2=8A+kI.$

Answer 1

Given,

$A=\left[\begin{array}{cc}1& 0\\ -1& 7\end{array}\right]$ and $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

To find $k$ such that $A^2=8A+kI$

Now, calculating $A^2$

$A^2=\left[\begin{array}{cc}1& 0\\ -1& 7\end{array}\right]\left[\begin{array}{cc}1& 0\\ -1& 7\end{array}\right]=\left[\begin{array}{cc}1\times 1+0\times -1& 1\times 0+0\times 7\\ -1\times 1+7\times -1& -1\times 0+7\times 7\end{array}\right]$

$\Rightarrow A^2=\left[\begin{array}{cc}1+0& 0+0\\ -1-7& 0+49\end{array}\right]=\left[\begin{array}{cc}1& 0\\ -8& 49\end{array}\right]$

Also, $\Rightarrow 8A+kI=8\left[\begin{array}{cc}1& 0\\ -1& 7\end{array}\right]+k\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

$\Rightarrow 8A+kI=\left[\begin{array}{cc}8\times 1& 0\times 8\\ -8\times -1& 7\times 8\end{array}\right]+\left[\begin{array}{cc}k& 0\\ 0& k\end{array}\right]$

$8A+kI=\left[\begin{array}{cc}8& 0\\ -8& 56\end{array}\right]+\left[\begin{array}{cc}k& 0\\ 0& k\end{array}\right]$

$\Rightarrow 8A+kI=\left[\begin{array}{cc}8+k& 0\\ -8& 56+k\end{array}\right]$

Given $A^2=8A+kI$

$\Rightarrow \left[\begin{array}{cc}1& 0\\ -8& 49\end{array}\right]=\left[\begin{array}{cc}8+k& 0\\ -8& 56+k\end{array}\right]$

Comparing each element

$1=8+k$

$\Rightarrow k=1-8$

$\Rightarrow k=-7$

Hence $k=-7.$